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I'm almost there with this one... I just can't make the final leap.
"Prove that if x,y,z are integers such that 9|(x^2+y^2+z^2), then 9|(x^2-y^2) or 9|(y^2-z^2) or 9|(z^2-x^2)."
The quadratic residues (mod 9) are 1, 4 and 7. Therefore the only ways that three squares can be summed to give 0 (mod 9) are (0,0,0); (1,1,7); (1,4,4); and (4,7,7) and any permutations of these triples. In all cases there is a repeat, but does this / why does this mean that 9|(x^2-y^2) or 9|(y^2-z^2) or 9|(z^2-x^2)?
I can sort of see it, but I can't explain it.
Thanks
Last edited by Daniel123 (2008-11-12 14:02:03)
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The fact that there's always a repeat is the key thing.
This means that there are always a and b in the triple so that a^2 = b^2 (mod 9).
So a^2 - b^2 = 0 (mod 9), ie. 9| (a^2 - b^2).
Now prove that {a,b} has to be one of {x,y}; {x,z} and {y,z} and you're there.
Why did the vector cross the road?
It wanted to be normal.
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thanks.
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